๐—–๐—ฆ ๐—”๐—น๐—ด๐—ผ ๐Ÿ’ป ๐ŸŒ ใ€Ž๐—–๐—ผ๐—บ๐—ฝ๐—ฒ๐˜๐—ถ๐˜๐—ถ๐˜ƒ๐—ฒ ๐—ฃ๐—ฟ๐—ผ๐—ด๐—ฟ๐—ฎ๐—บ๐—บ๐—ถ๐—ป๐—ดใ€
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๐ŸšฉMain Group - @SuperExams
๐Ÿ“Job Updates - @FresherEarth

๐Ÿ”ฐAuthentic Coding Solutions(with Outputs)
โš ๏ธDaily Job Updates
โš ๏ธHackathon Updates & Solutions

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from collections import defaultdict

def maxShared(employees_nodes, employees_from, employees_to, employees_weight):
    graph = defaultdict(lambda: defaultdict(int))

    for i in range(len(employees_from)):
        u, v, w = employees_from[i], employees_to[i], employees_weight[i]
        graph[u][v] += 1
        graph[v][u] += 1
   
    max_product = 0
   
    for i in range(1, employees_nodes + 1):
        for j in graph[i]:
            if graph[i][j] > 1: 
                max_product = max(max_product, i * j)
   
    return max_product


Movelsync Employee shared Interest โœ…
def solve(N, K):
    MOD = 10**9 + 7
    if N > 10*K:
        return 0

    if N == 10*K:
        return 1

    dp = [1] + [0] * N

    for _ in range(10):
        new_dp = [0] * (N + 1)
        for i in range(N + 1):
            for j in range(min(K, N - i) + 1):
                new_dp[i + j] = (new_dp[i + j] + dp[i]) % MOD
        dp = new_dp
   
    return dp[N]
๐—–๐—ฆ ๐—”๐—น๐—ด๐—ผ ๐Ÿ’ป ๐ŸŒ ใ€Ž๐—–๐—ผ๐—บ๐—ฝ๐—ฒ๐˜๐—ถ๐˜๐—ถ๐˜ƒ๐—ฒ ๐—ฃ๐—ฟ๐—ผ๐—ด๐—ฟ๐—ฎ๐—บ๐—บ๐—ถ๐—ป๐—ดใ€
Photo
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int maxi = 0;
int numWays = 0;

void backtrack(vector<int>& P, vector<vector<int>>& adj, vector<bool>& vis, int curr, int N) {
    bool all = true;
    for (int i = 0; i < N; ++i) {
        if (!vis[i]) {
            all = false;
            vis[i] = true;

            vector<int> prev;
            for (int nei : adj[i]) {
                if (!vis[nei]) {
                    vis[nei] = true;
                    prev.push_back(nei);
                }
            }

            backtrack(P, adj, vis, curr + P[i], N);

    
            for (int nei : prev) {
                vis[nei] = false;
            }
            vis[i] = false;
        }
    }

    if (all) {
        if (curr > maxi) {
            maxi = curr;
            numWays = 1;
        } else if (curr == maxi) {
            numWays++;
        }
    }
}

int main() {
    int N, M;
    cin >> N >> M;
    vector<int> P(N);
    for (int i = 0; i < N; ++i) {
        cin >> P[i];
    }

    vector<vector<int>> adj(N);
    for (int i = 0; i < M; ++i) {
        int x, y;
        cin >> x >> y;
        x--; y--;
        adj[x].push_back(y);
        adj[y].push_back(x);
    }

    vector<bool> vis(N, false);
    backtrack(P, adj, vis, 0, N);

    cout << maxi << " " << numWays << endl;

    return 0;
}

//phone pe
Cookie salesโœ…
Source : Hola
โค2
๐—–๐—ฆ ๐—”๐—น๐—ด๐—ผ ๐Ÿ’ป ๐ŸŒ ใ€Ž๐—–๐—ผ๐—บ๐—ฝ๐—ฒ๐˜๐—ถ๐˜๐—ถ๐˜ƒ๐—ฒ ๐—ฃ๐—ฟ๐—ผ๐—ด๐—ฟ๐—ฎ๐—บ๐—บ๐—ถ๐—ป๐—ดใ€
Photo
#include <iostream>
#include <vector>

using namespace std;

long long calc(int base, int times) {
    return base * (1LL << (times - 1));
}

int main() {
    int n, k;
    cin >> n >> k;

    vector<int> discounts(n);
    for (int i = 0; i < n; ++i) {
        cin >> discounts[i];
    }

   
    vector<int> times(n, 1);

  
    for (int i = 0; i < k; ++i) {
        long long max_increase = -1;
        int max_index = -1;
        for (int j = 0; j < n; ++j) {
            long long curr = calc(discounts[j], times[j]);
            long long next = calc(discounts[j], times[j] + 1);
            if (next - curr > max_increase) {
                max_increase = next - curr;
                max_index = j;
            }
        }
        times[max_index]++;
    }

  
    long long max_discount = 0;
    for (int i = 0; i < n; ++i) {
        max_discount |= calc(discounts[i], times[i]);
    }

    cout << max_discount << endl;

    return 0;
}

First trip of Luna โœ…
Phonepe
Source :Hola
๐—–๐—ฆ ๐—”๐—น๐—ด๐—ผ ๐Ÿ’ป ๐ŸŒ ใ€Ž๐—–๐—ผ๐—บ๐—ฝ๐—ฒ๐˜๐—ถ๐˜๐—ถ๐˜ƒ๐—ฒ ๐—ฃ๐—ฟ๐—ผ๐—ด๐—ฟ๐—ฎ๐—บ๐—บ๐—ถ๐—ป๐—ดใ€
Photo
#include <bits/stdc++.h>
#define ll long long
using namespace std;
struct TrieNode {
    unordered_map<ll, TrieNode*> children;
    ll count;

    TrieNode() : count(0) {}
};
class Trie {
public:
    Trie() {
        root = new TrieNode();
    }
    void insert(vector<ll>& vector)
{
        TrieNode* node = root;
        for (ll num : vector) {
            if (node->children.find(num) == node->children.end()) {
                node->children[num] = new TrieNode();
            }
            node = node->children[num];
            node->count++;
        }
    }
    ll countPrefix(vector<ll>& prefix) {
        TrieNode* node = root;
        for (ll num:prefix)
      {
           if(num==-1) continue;
            if (node->children.find(num) == node->children.end()) {
                return 0;
             }
            node = node->children[num];
       }
        return node->count;
    }
private:
    TrieNode* root;
};
signed main()
{
    ll n,m; cin>>n>>m;
vector<vector<ll>>a(n,vector<ll>(m));
for(ll i=0;i<n;i++)
{
  for(ll j=0;j<m;j++) cin>>a[i][j];
}
Trie trie;
    for(auto& vector:a) {
        trie.insert(vector);
    }
ll q; cin>>q;
vector<vector<ll>>qu(q);
for (ll i=0;i<q;i++)
{
        ll num;
        while(cin>>num && num!=-1)
  {
            qu[i].push_back(num);
        }
    }
for(auto&query:qu)
{
        cout<<trie.countPrefix(query)<<endl;
    }

    return 0;           
}
We're looking for for a full-time intern to work on a computer vision project in the Intelligent Interior team of Mercedes Benz R&D India (MBRDI).

Project duration: 6 months (on-site preferred)
Internship type: Full-time
Topic: 3D computer vision with prior experience in Gaze Estimation and/or Synthetic Data preferred
Minimum Qualification: Pursuing Bachelors/Post-graduation/PhD/Research in relevant disciplines
Preferred Qualification: Pursuing Post-graduation/PhD/Research at top-tier institutes in relevant disciplines

Please drop me an email with a detailed paragraph on your research experience along with your CV at : anuraag.bhattacharya98@gmail.com. DMs might not be entertained.

Background: We at MBRDI are working towards problems in Intelligent Car Interiors. The team is founded on a research mindset and has experience in publishing in some top tier conferences in Computer Vision and Machine learning. We adapt our research into innovative products used for cars.
โœ๏ธTECH MAHINDRA Interview Exp โœ…

1) Self Intro.
2) Project.
3) Qns on Project.
4) Which Programming Language
     you r Familiar with It.
5) What is Primarykey, Foreignkey.
6) Views in SQL.
7) Syntax of View.
8) Rank.
9) External Table.
10) Method Overloading.
11) Method Overriding.
12) Inheritance.
13) Explain Types of Inheritance
14) Why Multiple Inheritance is not
       Supported.
15) Is Multiple Inheritance.
      Supported injava and wha is it.
16) Relocation.
17) Night Shifts.
18) Why you Choose TechM.
19) About TechM.
๐—–๐—ฆ ๐—”๐—น๐—ด๐—ผ ๐Ÿ’ป ๐ŸŒ ใ€Ž๐—–๐—ผ๐—บ๐—ฝ๐—ฒ๐˜๐—ถ๐˜๐—ถ๐˜ƒ๐—ฒ ๐—ฃ๐—ฟ๐—ผ๐—ด๐—ฟ๐—ฎ๐—บ๐—บ๐—ถ๐—ป๐—ดใ€
Photo
long long minCost(vector<int>& a, vector<int>& b) {
        map<int,int>mp;
        int n=a.size();
        int mini=INT_MAX;
        for(int i=0;i<n;i++){
            mp[a[i]]++;
            mp[b[i]]--;
            mini=min(mini,a[i]);
            mini=min(mini,b[i]);
        }
        vector<int>x;
        for(auto it:mp){
            int t=it.second;
            if(t%2==1)return -1;
            else{
                for(int i=0;i<abs(t)/2;i++){
                    x.push_back(it.first);
                }
            }
        }
        long long ans=0;
        int m=x.size();
        for(int i=0;i<m/2;i++){
            ans+=min(x[i],2*mini);
        }
        return ans;
    }

//rearrange students โœ…
def count_subsequences(s1, s2):
    m, n = len(s1), len(s2)
    dp = [[0] * (n + 1) for _ in range(m + 1)]

    for j in range(n + 1):
        dp[0][j] = 1

    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if s1[i - 1] == s2[j - 1]:
                dp[i][j] = dp[i][j - 1] + dp[i - 1][j - 1]
            else:
                dp[i][j] = dp[i][j - 1]

    return dp[m][n]

//string subsequence
๐—–๐—ฆ ๐—”๐—น๐—ด๐—ผ ๐Ÿ’ป ๐ŸŒ ใ€Ž๐—–๐—ผ๐—บ๐—ฝ๐—ฒ๐˜๐—ถ๐˜๐—ถ๐˜ƒ๐—ฒ ๐—ฃ๐—ฟ๐—ผ๐—ด๐—ฟ๐—ฎ๐—บ๐—บ๐—ถ๐—ป๐—ดใ€
Photo
public static int pointsBelong(int x1, int y1, int x2, int y2, int x3, int y3, int xp, int yp, int xq, int yq) {
        if (!isValidTriangle(x1, y1, x2, y2, x3, y3)) {
            return 0;
        }

        boolean pBelongs = isPointInTriangle(x1, y1, x2, y2, x3, y3, xp, yp);
        boolean qBelongs = isPointInTriangle(x1, y1, x2, y2, x3, y3, xq, yq);

        if (pBelongs && qBelongs) {
            return 3;
        } else if (pBelongs) {
            return 1;
        } else if (qBelongs) {
            return 2;
        } else {
            return 4;
        }
    }

    private static boolean isValidTriangle(int x1, int y1, int x2, int y2, int x3, int y3) {
        double ab = distance(x1, y1, x2, y2);
        double bc = distance(x2, y2, x3, y3);
        double ac = distance(x1, y1, x3, y3);
        return (ab + bc > ac) && (bc + ac > ab) && (ac + ab > bc);
    }

    private static double distance(int x1, int y1, int x2, int y2) {
        return Math.sqrt(Math.pow(x2 - x1, 2) + Math.pow(y2 - y1, 2));
    }

    private static boolean isPointInTriangle(int x1, int y1, int x2, int y2, int x3, int y3, int xp, int yp) {
        double areaABC = triangleArea(x1, y1, x2, y2, x3, y3);
        double areaPAB = triangleArea(xp, yp, x1, y1, x2, y2);
        double areaPAC = triangleArea(xp, yp, x1, y1, x3, y3);
        double areaPBC = triangleArea(xp, yp, x2, y2, x3, y3);

        return Math.abs((areaPAB + areaPAC + areaPBC) - areaABC) < 1e-9;
    }

    private static double triangleArea(int x1, int y1, int x2, int y2, int x3, int y3) {
        return Math.abs(x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2)) / 2.0;
    }

Do they Belong
Oracle โœ…
๐Ÿ‘1